Noip2017Day1T3 逛公园

题目链接

problem

一个有向无重边自环图，设\(D\)为从\(1\)号点走到\(n\)号点的最短距离。问有多少条从\(1\)到\(n\)的路径长度不超过\(D+K\)。\(K\)为给定的值，且\(K\le 50\)

如果有无数条，输出-1

solution

下面有\(dis[i]\)表示\(i\)号点到\(n\)号点的最短路径长度。

设\(f[i][j]\)表示从\(i\)号点走到\(n\)号点，走了\(j\)的多余路径的方案数。就有如下转移:

\[f[i][j]=\sum\limits_{i,v之间有边}f[v][j-(dis[v]+w-dis[i])] \]

记忆化搜索即可。

注意到如果出现了无数条路径，肯定出现了0环。也就是某一个状态被访问了两次。记忆化搜索的过程中标记一下即可。

code

#include<cstdio>
#include<iostream>
#include<ctime>
#include<queue>
#include<cstring>
#include<string>
using namespace std;
typedef long long ll;
const int N = 200010;
ll read() {
	ll x = 0,f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		x = x * 10 + c - '0';
		c = getchar();
	}
	return x * f;
}
struct node {
	int v,nxt,w;
}e[N << 1],E[N << 1];
int head[N],ejs;
void add(int u,int v,int w) {
	e[++ejs].v = v;e[ejs].nxt = head[u];head[u] = ejs;e[ejs].w = w;
}
int head2[N],ejs2;
void add2(int u,int v,int w) {
	E[++ejs2].v = v;E[ejs2].w = w;E[ejs2].nxt = head2[u];head2[u] = ejs2;
}
queue<int>q;
int n,m,K,mod,dis[N],vis[N];
void spfa(int U) {
	memset(dis,0x3f,sizeof(dis));
	memset(vis,0,sizeof(vis));
	dis[U] = 0;
	q.push(U);
	while(!q.empty()) {
		int u = q.front();q.pop();vis[u] = 0;
		for(int i = head2[u];i;i = E[i].nxt) {
			int v = E[i].v;
			if(dis[v] > dis[u] + E[i].w) {
				dis[v] = dis[u] + E[i].w;
				if(!vis[v]) {
					vis[v] = 1;q.push(v);
				}
			}
		}
	}
}
int bz[N][60],f[N][60];
int dfs(int u,int x) {
	if(bz[u][x] == 2) return f[u][x];
	if(bz[u][x] == 1) return -1;
	bz[u][x] = 1;
	for(int i = head[u];i;i = e[i].nxt) {
		int v = e[i].v;
		int w = x - (dis[v] + e[i].w - dis[u]);
		if(w < 0 || w > K) continue;
		int k = dfs(v,w);
		if(k == -1) return -1;
		f[u][x] += k;
		f[u][x] %= mod;
	}
	bz[u][x] = 2;
	return f[u][x];
} 
int main() {
	int T = read();
	while(T--) {
		memset(head,0,sizeof(head));
		ejs2 = 0;
		memset(head2,0,sizeof(head2));
		ejs = 0;
		n = read(),m = read(),K = read(),mod = read();
		
		memset(f,0,sizeof(f));
		memset(bz,0,sizeof(bz));
		
		for(int i = 1;i <= m;++i) {
			int u = read(),v = read(),w = read();
			add(u,v,w);
			add2(v,u,w);
		}
		
		spfa(n);
		
		f[n][0] = 1;
		int ans = 0;
		for(int i = 0;i <= K;++i) {
			int k = dfs(1,i);
			if(k == -1) {
				ans = -1;break;
			}
			ans += k;
			ans %= mod;
		}
		printf("%d\n",ans);
	}
	
		
	return 0;
}